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Taking refrence to this question

One user is answering that -

In triangle ABC if $\vec{AA_1}$ median then

$\vec{AA_1} = \frac 12 (\vec{AB}+\vec{AC})$

I want to know Is this really a property?

If yes then what is its proof? Because I spent so much time to search but find nothing.

1 Answers 1

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enter image description here

Let $\vec{AK}=2\vec{AA_1}$ Then $ABKC$ are parallelogram. Thus $$2\vec{AA_1}=\vec{AK}=\vec{AB}+\vec{BK}=\vec{AB}+\vec{AC}$$ Then $$\vec{AA_1}=\frac{\vec{AB}+\vec{AC}}2$$

Addition:

enter image description here

In generally:

If $$\frac{AM}{MB}=\frac mn$$ then enter image description here

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    AK = AB+BK. How AB+BC? – 2017-01-12
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    No. It's $$|AK|=|AB|+BK$$ – 2017-01-12
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    Then why you write BC in step 2? – 2017-01-12
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    For any point $B$ (may be $B \not \in AC$) $$\vec{AB}+\vec{BC}=\vec{AC}$$ – 2017-01-12
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    You are not understanding. AK = AB + BK. But you are writing AK = AB + BC. – 2017-01-12
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    And I know addition property. – 2017-01-12
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    Sorry! I edited! – 2017-01-12
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    One last question if two vectors are equal we can say that it is parallel? – 2017-01-13
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    @JohnSr: Yes!!!!! – 2017-01-13